Further Mathematics Higher Level - IB Documents
Further mathematicsHigher levelSpecimen papers 1 and 2For first examinations in 2014
CONTENTSFurther mathematics higher level paper 1 specimen question paperFurther mathematics higher level paper 1 specimen markschemeFurther mathematics higher level paper 2 specimen question paperFurther mathematics higher level paper 2 specimen markscheme
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–2–SPEC/5/FURMA/HP1/ENG/TZX/XXFull marks are not necessarily awarded for a correct answer with no working. Answers must be supportedby working and/or explanations. In particular, solutions found from a graphic display calculator should beyour answer. Where an answer is incorrect, some marks may be given for a correct method, provided thisis shown by written working. You are therefore advised to show all working.1.[Maximum mark: 6]Using l’Hôpital’s Rule, determine the value oft x x . x s x2.[Maximum mark: 8](a)Show that the following vectors form a basis for the vector space3. (b)Express the following vector as a linear combination of the above vectors. 3.[3 marks][5 marks][Maximum mark: 10]The positive integer N is represented by 4064 in base b and 2612 in base b 1.(a)Determine the value of b .(b)Find the representation of N(i)in base 10;(ii)in base 12.[4 marks][6 marks]
–3–4.SPEC/5/FURMA/HP1/ENG/TZX/XX[Maximum mark: 11]The weights of potatoes in a shop are normally distributed with mean 98 grams andstandard deviation 16 grams.(a)(b)5.The shopkeeper places 100 randomly chosen potatoes on a weighing machine.Find the probability that their total weight exceeds 10 kilograms.[3 marks]Find the minimum number of randomly selected potatoes which are needed toensure that their total weight exceeds 10 kilograms with probability greaterthan 0.95.[8 marks][Maximum mark: 11](a)The point lies on the parabolathe parabola at T has equation t② ①. Show that the tangent to.[3 marks]t(b)The distinct points P and , where q , also lie onthe parabola . Given that the line (PQ) passes through the focus, showthat(i)(ii)6. 1;the tangents to the parabola at P and Q, intersect on the directrix.[Maximum mark: 7]The group ④G ⑥The group H Prove that G and H are isomorphic.7.[8 marks]. ⑤ [Maximum mark: 9](a)Given that (b)Show that❲❳❨❩ ❯❱, show that is exactly divisible by 7.for all n .[4 marks][5 marks]Turn over
–4–8.[Maximum mark: 8] 9.SPEC/5/FURMA/HP1/ENG/TZX/XX s .(a)Show that the series is conditionally convergent but not absolutely convergent.[6 marks](b)Show that S[2 marks] .[Maximum mark: 8]Consider the system of equations (a) ① ② ③ .By reducing the augmented matrix to row echelon form,k for which the system has a solution.(b)10.For this value of k , determine the solution.[5 marks][3 marks][Maximum mark: 11]Bill is investigating whether or not there is a positive association between the heights ,whereof boys of this age. He measures the height, h cm, and weight, w kg, of each of arandom sample of 20 boys of this age and he calculates the following statistics. (This question continues on the following page)
–5–SPEC/5/FURMA/HP1/ENG/TZX/XX(Question 10 continued)(ii)Calculate the p-value of your result and interpret it at the 1 % level of[8 marks](b)11.(i)Calculate the equation of the least squares regression line of w on h .(ii)The height of a randomly selected boy of this age of 90 cm. Estimate hisweight.[Maximum mark: 11]The function f12.[3 marks] ① s ①.[2 marks].(a)Show that(b)Determine the Maclaurin series for(c)By differentiating your series, determine the Maclaurin series forthe term in x3. up to and including the term in x 4.x [5 marks]x up to[4 marks][Maximum mark: 14]The matrix A is given byA(a)(b) .(i)Find the matrices A2 and A3 , and verify that A A A.(ii)Deduce that A A A.(i)Suggest a similar expression for An in terms of A and A2 , valid for n 3.(ii)Use mathematical induction to prove the validity of your suggestion.[6 marks][8 marks]Turn over
–6–13.SPEC/5/FURMA/HP1/ENG/TZX/XX[Maximum mark: 9]A sequence ④ ⑥ , n . Obtain anexpression for in terms of n given that u0 0 and u1 1.14.[Maximum mark: 13]The set S contains the eight matrices of the form 0 0 0 0 0 0 where a , b , c can each take one of the values 1 or 1.15.(a)Show that any matrix of this form is its own inverse.[3 marks](b)Show that S forms an Abelian group under matrix multiplication.[9 marks](c)Giving a reason, state whether or not this group is cyclic.[1 mark][Maximum mark: 14](a)(b)Prove the internal angle bisector theorem, namely that the internal bisector of anangle of a triangle divides the side opposite the angle into segments proportionalto the sides adjacent to the angle.[6 marks]The bisector of the exterior angle A of the triangle ABC meets (BC) at P.The bisector of the interior angle B meets [AC] at Q. Given that (PQ) meets .[AB] at R, use Menelaus’ theorem to prove that (CR) bisects the angle [8 marks]
THER MATHEMATICSHigher LevelPaper 116 pages
–2–SPEC/5/FURMA/HP1/ENG/TZX/XX/MInstructions to ExaminersAbbreviationsMMarks awarded for attempting to use a correct Method; working must be seen.(M)Marks awarded for Method; may be implied by correct subsequent working.AMarks awarded for an Answer or for Accuracy; often dependent on preceding M marks.(A)Marks awarded for an Answer or for Accuracy; may be implied by correct subsequent working.RMarks awarded for clear Reasoning.NMarks awarded for correct answers if no working shown.AGAnswer given in the question and so no marks are awarded.Using the markscheme1GeneralWrite the marks in red on candidates’ scripts, in the right hand margin. Show the breakdown of individual marks awarded using the abbreviations M1, A1, etc. Write down the total for each question (at the end of the question) and circle it.2Method and Answer/Accuracy marks Do not automatically award full marks for a correct answer; all working must be checked, and 3marks awarded according to the markscheme.It is not possible to award M0 followed by A1, as A mark(s) are often dependent on the precedingM mark.Where M and A marks are noted on the same line, e.g. M1A1, this usually means M1 for anattempt to use an appropriate method (e.g. substitution into a formula) and A1 for using thecorrect values.Where the markscheme specifies (M2), N3, etc. do not split the marks.Once a correct answer to a question or part-question is seen, ignore further working.N marksAward N marks for correct answers where there is no working. Do not award a mixture of N and other marks. There may be fewer N marks available than the total of M, A and R marks; this is deliberate as itpenalizes candidates for not following the instruction to show their working.
–3–4SPEC/5/FURMA/HP1/ENG/TZX/XX/MImplied marksImplied marks appear in brackets e.g. (M1), and can only be awarded if correct work is seen or ifimplied in subsequent working. Normally the correct work is seen or implied in the next line. Marks without brackets can only be awarded for work that is seen.5Follow through marksFollow through (FT) marks are awarded where an incorrect answer from one part of a question isused correctly in subsequent part(s). To award FT marks, there must be working present and notjust a final answer based on an incorrect answer to a previous part. If the question becomes much simpler because of an error then use discretion to award fewer FTmarks. If the error leads to an inappropriate value (e.g. sin 1.5 ), do not award the mark(s) for the finalanswer(s). Within a question part, once an error is made, no further dependent A marks can be awarded, butM marks may be awarded if appropriate. Exceptions to this rule will be explicitly noted on the markscheme.6Mis-readIf a candidate incorrectly copies information from the question, this is a mis-read (MR). Apply a MRpenalty of 1 mark to that question. Award the marks as usual and then write –1(MR) next to the total.Subtract 1 mark from the total for the question. A candidate should be penalized only once for aparticular mis-read. If the question becomes much simpler because of the MR, then use discretion to award fewermarks. If the MR leads to an inappropriate value (e.g. sin 1.5 ), do not award the mark(s) for thefinal answer(s).7Discretionary marks (d)An examiner uses discretion to award a mark on the rare occasions when the markscheme does notcover the work seen. The mark should be labelled (d) and a brief note written next to the markexplaining this decision.8Alternative methodsCandidates will sometimes use methods other than those in the markscheme. Unless the questionspecifies a method, other correct methods should be marked in line with the markscheme. If in doubt,contact your team leader for advice. Alternative methods for complete questions are indicated by METHOD 1, METHOD 2, etc. Alternative solutions for part-questions are indicated by EITHER . . . OR. Where possible, alignment will also be used to assist examiners in identifying where thesealternatives start and finish.
–4–9SPEC/5/FURMA/HP1/ENG/TZX/XX/MAlternative formsUnless the question specifies otherwise, accept equivalent forms. As this is an international examination, accept all alternative forms of notation.In the markscheme, equivalent numerical and algebraic forms will generally be written inbrackets immediately following the answer.In the markscheme, simplified answers, (which candidates may not write in examinations), willgenerally appear in brackets. Marks should be awarded for either the form preceding the bracket orthe form in brackets (if it is seen).Example: for differentiating f ( x) 2sin (5 x 3) , the markscheme gives:f ( x) 2 cos (5 x 3) 5 10cos (5 x 3) A1Award A1 for 2cos (5 x 3) 5 , even if 10cos (5 x 3) is not seen.10Accuracy of AnswersThe method of dealing with accuracy errors on a whole paper basis by means of the Accuracy Penalty(AP) no longer applies.Instructions to examiners about such numerical issues will be provided on a question by questionbasis within the framework of mathematical correctness, numerical understanding and contextualappropriateness.The rubric on the front page of each question paper is given for the guidance of candidates.The markscheme (MS) may contain instructions to examiners in the form of “Accept answers whichround to n significant figures (sf)”. Where candidates state answers, required by the question, tofewer than n sf, award A0. Some intermediate numerical answers may be required by the MS but notby the question. In these cases only award the mark(s) if the candidate states the answer exactly or toat least 2sf.11Crossed out workIf a candidate has drawn a line through work on their examination script, or in some other waycrossed out their work, do not award any marks for that work.12CalculatorsA GDC is required for paper 3, but calculators with symbolic manipulation features (e.g. TI-89) arenot allowed.Calculator notationThe Mathematics HL guide says:Students must always use correct mathematical notation, not calculator notation.Do not accept final answers written using calculator notation. However, do not penalize the use ofcalculator notation in the working.13More than one solutionWhere a candidate offers two or more different answers to the same question, an examiner should onlymark the first response unless the candidate indicates otherwise.
–5–1.tan x xsec 2 x 1 limx 0 1 cos xx 0sin x0this still repeat the process2sec2 x tan x limx 0cos x0M1A1A1ORtan 2 x limx 0 sin xsin x lim 2x 0 cos x0M1A1A1[6 marks]
–6–2.(a)SPEC/5/FURMA/HP1/ENG/TZX/XX/M1 2 5(M1)A1let A 2 3 2 and consider det ( A) 303 1 5the vectors form a basis because the determinant is non-zero (or becausethe matrix is non-singular)R1[3 marks](b) 12 1 2 5 let 14 2 3 2 16 3 1 5 M1A1so thatEITHER 2 5 122 3 2 143 5 16M1OR 1 2 5 12 2 3 2 14 3 1 5 16 M1THENgiving 3, 2, 1 12 1 2 5 hence 14 3 2 2 3 1 2 16 3 1 5 (A1)A1[5 marks]Total [8 marks]
–7–3.(a)SPEC/5/FURMA/HP1/ENG/TZX/XX/Mthe equation satisfied by b is4b3 6b 4 2(b 1)3 6(b 1) 2 (b 1) 2M1A12b 12b 13b 7 0b 732(A1)A1[4 marks](b)(i)(ii)N 4 73 6 7 4 1418or N 2 83 6 82 1 8 2 1418(M1)12 141812 11812 9(M1)A1remainder 2(A1)remainder A (where A 10 )(A1)1418 (9A2)12Note:A1Accept alternative correct methods.[6 marks]Total [10 marks]4.(a)let T denote the total weight, thenT N (9800, 25600)P (T 10000) 0.106(M1)(A1)A1[3 marks](b)let there be n potatoes, in this case,T N (98n , 256n)we requireP (T 10000) 0.95or equivalentlyP (T 10000) 0.05standardizing,10000 98n P Z 0.0516 n 10000 98n 1.6449 16 nA1(M1)98n 26.32 n 10000 0solving the corresponding equation, n 104.7 the required minimum value is 105Note:Part (b) could also be solved using SOLVER and normalcdf, or by trial andimprovement.Note:Allow the use ofA1A1(A1)A1(A1)A1instead of and throughout.[8 marks]Total [11 marks]
–8–5.(a)dy 4adxdy 2 a 1dxy tM12yNote:SPEC/5/FURMA/HP1/ENG/TZX/XX/MA1Accept parametric differentiation.the equation of the tangent is1y 2at ( x at 2 )txy attA1AGAccept equivalent based on y mx c .Note:[3 marks](b)(i)the focus F is (a , 0)A1EITHERthe gradient of (PQ) is2a ( p q )2 22a(p q ) p qthe equation of (PQ) is ysubstitute x a , y 0pq 12x2apq p q p qM1A1A1M1AGORthe condition for PFQ to be collinear is2a ( p q )2ap a ( p 2 q 2 ) ap 2 a22pp q p2 1p 2 1 p 2 pqpq 1Note:(ii)M1A1A1A1AGThere are alternative approaches.the equations of the tangents at P and Q arexxy ap and y aqpqthe tangents meet wherexx ap aqpqx apq athe equation of the directrix is x aso that the tangents meet on the directrixM1A1R1AG[8 marks]Total [11 marks]
–9–6.SPEC/5/FURMA/HP1/ENG/TZX/XX/Mconsider the function f : G H defined by f ( g ) 2 g where g Ggiven g1 , g 2 G , f ( g1 ) f ( g 2 ) 2 g1 2 g 2 g1 g 2 (injective)given h H then h 4n , so f (2n) h and 2n G (surjective)hence f is a bijectionthen, for g1 , g 2 Gf ( g1 g 2 ) 2( g1 g 2 )f ( g1 ) f ( g 2 ) 2 g1 2 g 2it follows that f ( g1 g 2 ) f ( g1 ) f ( g 2 )which completes the proof that {G , } and {H , } are isomorphicA1M1M1A1A1A1R1AG[7 marks]7.(a)a b (mod p ) a b pN , N (b pN )n b n nbn 1 pN b n pM where M this shows that a n b n (mod p )anM1M1A1A1AG[4 marks](b)29 1(mod 7) 2913 113 1(mod 7)13 1(mod 7) 13 ( 1) 1(mod 7)therefore 29 13 1 ( 1) 0(mod 7)this shows that 2913 1329 is exactly divisible by 729132929M1A1A1M1A1AG[5 marks]Total [9 marks]8.(a) 1 decreases as n increases n 1 sin 0 as n n sin A1A1so using the alternating series test, the series is conditionally convergentR1comparing (the absolute series) with the harmonic series: 1M1 nn 1 1 sin lim n 1A1n 1nsince the harmonic series is divergent, it follows by the limit comparisontheorem that the given series is not absolutely convergentR1hence the series is conditionally convergentAG[6 marks]continued
– 10 –SPEC/5/FURMA/HP1/ENG/TZX/XX/MQuestion 8 continued(b)successive partial sums are0.841 0.362 0.689 0.441 A1since S lies between any pair of successive partial sums, it follows that S liesbetween 0.441 and 0.689 R1and is therefore greater than 0.4AGNote:Use of the facts that the error is always less than the modulus of the nextterm, or the sequence of even partial sums gives lower bounds are equallyacceptable.[2 marks]Total [8 marks]9.(a)reducing to row echelon form1 1 25 70 4 50 8 10 140 4 5 k 15(M1)(A1)1 1 250 4 5 70 0 000 0 0 k 8A1(i)this shows that the rank of the matrix is 2A1(ii)the equations can be solved if kA18[5 marks](b)let z then y A15 74and x 5 2 Note:A15 74 13 3 4A1Accept equivalent expressions.[3 marks]Total [8 marks]
– 11 –34150 340 (M1)(A1) 3402 20022 5830 201124 20 20 Accept equivalent formula. 0.610A1 n 2 T R 1 R 2 has the t -distribution with n 2 degrees of freedom (ii)t 0.6097666 181 0.6097666 2M1 3.2640 DF 18p-value 0.00215 A1A1A1this is less than 0.01, so we conclude that there is a positive associationbetween heights and weights of boys of this ageR1[8 marks](b)(i)(ii)Note:the equation of the regression line of w on h is340 20 34150 340 2002 2002 w h 220 20 201124 2002 20 w 0.160h 0.957M1putting h 90 , w 15.4 (kg)A1A1Award M0A0A0 for calculation of h on w.[3 marks]Total [11 marks]11.f ( x) e x cos x e x sin x(a)A1f ( x) e cos x e sin x e sin x e cos xxxxxA1 2e sin xxAG[2 marks]f ( x) 2e x sin x 2e x cos x(b)fIV( x) 4e cos xf (0) 1, f (0) 1, f (0) 0, f (0) 2, fthe Maclaurin series isx3 x4e x cos x 1 x 3 6Note:A1xA1IV(0) 4(A1)M1A1Accept multiplication of series method.[5 marks]continued
– 12 –SPEC/5/FURMA/HP1/ENG/TZX/XX/MQuestion 11 continued(c)differentiating, 23x x2xe sin x 1 x 1 x 33 x x x3 3e x cos x e x sin x 1 x 23M1A132xM132A1[4 marks]Total [11 marks]12.(a)(i) 2 4 1 A 472 14 26 7 4 7 2 A 6103 24 41 12 2 4 1 0 12 A A 2 472 24 14 26 7 4 11 4 7 2 6 10 3 A 24 41 12 A AA A (2 A A) 2A A 2(2 A A) A 3A 2 A2A13A123(ii)4323Note:M1AGM1A1222 2 012A1AGAccept alternative solutions that include correct calculation of bothsides of the expression.[6 marks]continued
– 13 –SPEC/5/FURMA/HP1/ENG/TZX/XX/MQuestion 12 continued(b)(i)conjecture: An (n 1) A2 (n 2) A(ii)first check that the result is true for n 3the formula gives A3 2 A2 A which is correctassume the result for n k , i.e.Ak (k 1) A2 (k 2) AsoAk 1 A[(k 1) A2 (k 2) A] (k 1) A (k 2) A (k 1) (2 A2 A) (k 2) A2 kA2 (k 1) Aso true for n k true for n k 1 and since true for n 3 ,32the result is proved by inductionNote:A1A1M1M1A1M1A1R1Only award the R1 mark if a reasonable attempt at a proof byinduction has been made.[8 marks]Total [14 marks]13.the auxiliary equation is m 2 2m 5 0the roots are 1 2ithe general solution is un A (1 2i)n B (1 2i) nsubstituting u0 0 and u1 1A B 0A (1 2i) B (1 2i) 1Note:M1A1A1M1A1Only award above A1 if both equations are correct.solvingA (1 2i 1 2i) 11iA or 4i41iB or4i4therefore11iiun (1 2i)n (1 2i)n or (1 2i)n (1 2i)n4i4i44(M1)A1A1A1[9 marks]
– 14 –14.(a) a 0 0
Further mathematics higher level paper 1 specimen markscheme Further mathematics higher level paper 2 specimen question paper Further mathematics higher level paper 2 specimen markscheme . X X X A S F U R T H E R T T H I E C M M H I G H E R L E V E L P E P R S S O S O A D S D A D o o
2. Further mathematics is designed for students with an enthusiasm for mathematics, many of whom will go on to degrees in mathematics, engineering, the sciences and economics. 3. The qualification is both deeper and broader than A level mathematics. AS and A level further mathematics build from GCSE level and AS and A level mathematics.
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The proportion of students taking Mathematics (AS/A level) who are girls is around 40% with the corresponding figure being 30% for Further Mathematics (AS/A level). The Further Mathematics Support Programme (FMSP) promotes participation in Advanced level . Mathematics to all students who would benefit from taking the qualifications, especially .
A Level Further Mathematics A . Y540 Pure Core 1 . Sample Question Paper . Date – Morning/Afternoon. Time allowed: 1 hour 30 minutes . OCR supplied materials: Printed Answer Booklet Formulae A Level Further Mathematics A . You must have: Printed Answer Booklet Formulae A Level Further Mathematics A Scientific or graphical .
Cambridge International AS Level Further Mathematics makes up the first half of the Cambridge International A Level course in further mathematics and provides a foundation for the study of further mathematics at Cambridge International A Level. Depending on local university entrance requirements, students may be able to use
A: Further Pure Mathematics 2 B: Further Statistics 1 C: Further Mechanics 1 D: Decision Mathematics 1 E: Further Statistics 2 F: Further Mechanics 2 G: Decision Mathematics 2 Assessment overview Students must answer
Paper 2: Further Mathematics Options, Further Mechanics 1, Section 2.1 – Further guidance added 23 Paper 2: Further Mathematics Options, Further Mechanics 2, Section 3.1 – Text in the . Maths and Further Maths textbooks retain all the features you know and love about the current series, whilst being fully updated to match the new .File Size: 920KB
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